Áp dụng BĐT cauchy - schwer dưới dạng engel ta có :
\(\dfrac{x^2}{3}+\dfrac{y^2}{3}+\dfrac{z^2}{3}\ge\dfrac{\left(x+y+z\right)^2}{9}\)
\(\Leftrightarrow\) \(\dfrac{x^2+y^2+z^2}{3}\ge\left(\dfrac{x+y+z}{3}\right)^2\)
\(\dfrac{x^2+y^2+z^2}{3}\ge\left(\dfrac{x+y+z}{3}\right)^2\left(1\right)\\ \Leftrightarrow\dfrac{3\left(x^2+y^2+z^2\right)}{9}-\dfrac{\left(x+y+z\right)^2}{9}\ge0\\ \Leftrightarrow\dfrac{3x^2+3y^2+3z^2-x^2-y^2-z^2-2xy-2yz-2xz}{9}\ge0\\ \Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2xz\ge0\\ \Leftrightarrow\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(x^2-2xz+z^2\right)\ge0\\ \Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2\ge0\left(2\right)\)
Bất đẳng thức (2) luôn đúng \(\forall x\)
nên bất đẳng thức (1) luôn đúng.
Vậy \(\dfrac{x^2+y^2+z^2}{3}\ge\left(\dfrac{x+y+z}{3}\right)^2\)
đẳng thức xảy ra khi: \(x=y=z\)